---
title: "Surge numerical: 4 m tunnel, 4000 m, 20 m³/s"
subject: "Hydropower Engineering"
chapter: 6
key: "semester_7/hydropower_engineering:surge-numerical-tunnel-4m-4000m-20cumec"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/surge-numerical-tunnel-4m-4000m-20cumec
site: BCT Engineering
---

# Surge numerical: 4 m tunnel, 4000 m, 20 m³/s

Hydropower Engineering · Chapter 6: Water Conveyance Structures

- Asked in: 2080 Baishakh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/surge-numerical-tunnel-4m-4000m-20cumec
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-6

## Question

In a hydropower project the headrace tunnel of 4.0 m diameter and 4000 m length carries 20 m³/s discharge to the surge tank of 10 m diameter. The penstock from the surge tank to the power house has 3.2 m diameter and 700 m length. Considering the case of instantaneous closure, find the maximum height of the surge tank required and the time period of oscillation of wave. Assume friction factor = 0.018.

## Short answer

Tunnel velocity 1.592 m/s, friction loss 2.32 m and frictionless amplitude 12.86 m. With instantaneous closure the maximum upsurge is 11.36 m above reservoir level; first downswing is 9.36 m below. Oscillation period is 317.2 s, about 5.3 min.

## Answer

#### Method used
Let the tunnel have area $A_t$, length $L$, diameter $D$ and friction factor $f$; the tank has area $A_s$.

- Tunnel velocity $V = Q/A_t$; steady friction loss $h_f = \dfrac{fLV^2}{2gD}$.
- Frictionless amplitude: $Z_0 = V\sqrt{\dfrac{L A_t}{g A_s}}$.
- Upsurge with friction (full load rejection), $Z$ measured above reservoir level, from the equation of motion and continuity:

$$
\frac{Z_{max}}{h_f}=\frac{Z_0^2}{2h_f^2}\left[1-e^{-2h_f (Z_{max}+h_f)/Z_0^2}\right]
$$

- The following down-surge is found by integrating the reversed-flow equations step by step (below reservoir level).
- Period of mass oscillation: $T = 2\pi\sqrt{\dfrac{L A_s}{g A_t}}$.

#### Calculation
$A_t = \frac{\pi}{4}(4)^2 = 12.566\ \text{m}^2$, $A_s = \frac{\pi}{4}(10)^2 = 78.54\ \text{m}^2$, $L = 4000$ m, $f = 0.018$.

- $V = 20/12.566 = 1.592$ m/s
- $h_f = \dfrac{0.018 \times 4000 \times 1.592^2}{2(9.81)(4)} = 2.32$ m
- $Z_0 = 1.592\sqrt{\dfrac{4000 \times 12.566}{9.81 \times 78.54}} = 12.86$ m

Instantaneous closure of the turbine gates means complete rejection of the discharge, so the water rises in the tank. The penstock (3.2 m, 700 m) lies downstream of the tank, so it does not affect the tank oscillation.

#### Maximum height of surge tank
From the upsurge equation with friction: $Z_{max} = 11.36$ m above reservoir level (the frictionless value is 12.86 m).

The tank must therefore extend at least 11.4 m above the reservoir water level (plus a freeboard of about 1 to 2 m). The first downswing is 9.36 m below reservoir level.

#### Time period

$$
T = 2\pi\sqrt{\frac{4000 \times 78.54}{9.81 \times 12.566}} = 317.2\ \text{s} \approx 5.3\ \text{min}
$$

**Answer: maximum upsurge height $= 11.36$ m above reservoir level; $T = 317.2$ s.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the tunnel velocity for 20 m3/s through a 4 m diameter tunnel?
   A. About 1.59 m/s
   B. About 0.40 m/s
   C. About 6.37 m/s
   D. About 5.0 m/s
   Answer: A. V = Q/At = 20/12.566 = 1.592 m/s. (Key point: Tunnel velocity)

2. What happens to the tank level on instantaneous closure of the turbine gates?
   A. Water falls first to the maximum downsurge
   B. Flow is completely rejected, so water rises in the tank to the maximum upsurge
   C. The level stays unchanged
   D. Water hammer travels up the tunnel
   Answer: B. Full rejection forces tunnel flow into the tank, raising its level above reservoir level. (Key point: Instantaneous closure)

3. What minimum height above reservoir level must the tank have, and how does it relate to the frictionless value?
   A. 12.86 m exactly with no freeboard
   B. 2.32 m, equal to the friction loss
   C. At least 11.4 m (with friction) which is less than the frictionless 12.86 m, plus freeboard
   D. 9.36 m, the downswing value
   Answer: C. With friction Zmax = 11.36 m, below Z0 = 12.86 m; add 1 to 2 m freeboard. (Key point: Maximum height of tank)

4. Which period results for this tank?
   A. T = 12.86 s
   B. T = 158.6 s
   C. T = 700 s
   D. T = 317.2 s
   Answer: D. T = 2 pi sqrt(4000 x 78.54 / (9.81 x 12.566)) = 317.2 s. The penstock does not affect it. (Key point: Period)

## Close variants set in other exams

- [Surge numerical: 4.5 m tunnel, 2500 m, 25 m³/s](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-tunnel-4-5m-2500m-25cumec) — Hydropower Engineering, chapter 6, asked 2×

## Related in Hydropower Engineering

- [Surge numerical: 13 m tank, 1.8 km tunnel](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-13m-tank-1-8km-tunnel) — Hydropower Engineering, chapter 6
- [Surge and water hammer: 45 m³/s, 20 m surge tank](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-45cumec-3-penstocks) — Hydropower Engineering, chapter 6, asked 2×
- [Surge tank: tunnel 5 m dia, 28 cumec](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-tunnel-5m-28cumec) — Hydropower Engineering, chapter 6
- [Surge and water hammer: 60 m³/s, 30 m surge tank](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-60cumec-3-penstocks-30m-tank) — Hydropower Engineering, chapter 6, asked 2×
- [Surge and water hammer: 25 m³/s, 15 m surge tank](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-25cumec-2-penstocks-15m-tank) — Hydropower Engineering, chapter 6, asked 3×
- [Surge tank: 7.5 m³/s through 2.5 m tunnel](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-7-5cumec-2-5m) — Hydropower Engineering, chapter 6

## Related in other subjects

- [Surge tank objectives and economic penstock diameter factors](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asurge-tank-objectives-economic-diameter) — Hydro Power, chapter 5
- [Water hammer in penstock and surge tank](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Awater-hammer-surge-tank) — Power Plant Equipment, chapter 1, asked 2×
- [Pressure rise for instantaneous closure, elastic pipe](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aelastic-pipe-pressure-rise-derivation) — Hydraulics, chapter 4